@draks Well, I know the proof that $\sin(\mathbb N)$ is dense. Actually, I just remembered it from my first course in analysis and thought about this problem. The proof of the case with $\mathbb N$ doesn't seem to generalize, and it would be strange if it did I think. But I have simply no idea how to find another approach.
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user23211Feb 15 '12 at 22:57

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I don't know the $\sin(\mathbb{N})$ proof (can you provide a link?), but would it help to think of $\mathbb{N}$ as sum of all primes, semi-primes, k-almost primes?
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draks ...Feb 15 '12 at 23:09

@draks There's a question about it on this site. There's a link to a paper with a proof there but I can't access it from my house so I'm not sure what's in it.
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user23211Feb 15 '12 at 23:12

1 Answer
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The sequence of all multiples of an irrational $\alpha$ by successive prime
numbers, $2 \alpha$, $3 \alpha$, $5 \alpha$, $7 \alpha$, $11 \alpha$, ... is
equidistributed modulo 1. This is a famous theorem of analytic number
theory, proved by I. M. Vinogradov in 1935.

With $\alpha = \frac{1}{2 \pi}$, this implies that $P$ is equidistributed modulo $2 \pi$. Using this, and the continuity of the sine function, I think it is straightforward to show that $\sin(P)$ is dense in $[-1,1]$ (although not equidistributed).

Also, the distribution of $sin(P)$ can be derived from the fact of the equidistribution of $P$ modulo $2\pi$.
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Michael LugoFeb 16 '12 at 5:37

Dan, could you post a link or a reference to Vinogradov's proof?
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user23211Feb 16 '12 at 8:11

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@ymar, I couldn't find one with a search, but here it is claimed to be "a byproduct of [Vinogradov's work on] the odd Goldbach conjecture". It is said to have been proven in 1935 so probably the document can be narrowed down on that basis.
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Dan BrumleveFeb 16 '12 at 8:34

How about $P^{it}$? Is this dense over the unit circle?
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draks ...Jul 16 '12 at 14:40

@draks The phase of $P^{it}$ varies extremely slowly for any fixed $t$, so there's no question that it is dense on the unit circle. One only needs $\frac{p_{n+1}}{p_n} \to 1$, which is slightly stronger than Bertrand's postulate and weaker than the Prime Number Theorem.
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Erick WongJan 6 '13 at 7:40